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Poisson's equation and Green's functions

  1. Nov 26, 2007 #1

    jmb

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    Trying to verify the general integral solution to Poisson's equation I reach the following contradiction, what am I missing/doing wrong?

    A solution [tex]u(\mathbf{x})[/tex] to Poisson's equation satisfies:

    [tex]\nabla^2 u(\mathbf{x}) = - \frac{\rho(\mathbf{x})}{\epsilon_0}[/tex]

    we can find such a solution in a given domain by evaluating:

    [tex]u(\mathbf{x}) = \int_{V} \frac{\rho(\mathbf{x'})}{4 \pi \epsilon_0 |\mathbf{x}-\mathbf{x'}|} d\mathbf{x'}[/tex]

    or equivalently (if we are only interested in the electric field):

    [tex]\mathbf{E} = -\nabla u = - \int_{V} \frac{\rho(\mathbf{x'})}{4 \pi \epsilon_0} \frac{\mathbf{x}-\mathbf{x'}}{|\mathbf{x}-\mathbf{x'}|^3} d\mathbf{x'}[/tex]

    We can think of these solutions as either the Green's function solution or, physically, as the convolution of the point source solution with the source density [tex]\rho(\mathbf{x})[/tex].

    However if I take the Laplacian (or the divergence in the case of the electric field solution) of either of the above I get zero instead of [tex]-\frac{\rho}{\epsilon_0}[/tex]. What am I missing??

    To evaluate the Laplacian/divergence of the above I make use of the fact that [tex]\mathbf{x'}[/tex] is a 'dummy variable', which allows me to switch the order of integration and differentiation when applying [tex]\nabla^2[/tex] to the integral on the RHS and also means the components of [tex]\mathbf{x'}[/tex] all have derivative zero. With these assumptions the operation is fairly straightforward to carry out, but I have anyway verified my answer with Mathematica.
     
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  3. Nov 26, 2007 #2

    marcusl

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    Not so. You should take the Laplacian with respect to x (after all you are calculating [tex]\nabla^2{u(x)}[/tex]). Since
    [tex]\nabla^2 \frac{1}{|x-x'|}=\delta(x-x')[/tex]
    you get the right result from the integration.
    This expression is derived in many places:
    Jackson, Classical Electrodynamics, 2nd ed., Eq. (1.31)
    Arfken, Math Methods for Physicists, 2nd ed., sect. 1.15
    to give two examples.
     
    Last edited: Nov 26, 2007
  4. Nov 26, 2007 #3

    jmb

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    Thanks marcusl.

    Yes I was taking the derivative wrt x (hence the derivatives of all components of x' being zero). But I was erroneously evaluating the Laplacian of [tex]\frac{1}{|x-x'|}[/tex] as zero.

    I've since found references to the result you quote, but they all start by searching for the Green's function [tex]G[/tex] that satisfies [tex]\nabla^2 G = \delta(x-x')[/tex] and then deducing it to be the above... out of interest do you know of any way to show it starting from the other end --- i.e. directly differentiating [tex]\frac{1}{|x-x'|}[/tex] and showing its Laplacian to be the delta function? Or do the references you quote do it that way?

    I'll look them up when I get the chance. Thanks again.
     
  5. Nov 26, 2007 #4

    marcusl

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    Yes, Jackson shows it explicitly. And, now that I'm looking at his formula instead of relying on memory, I see I forgot the constant
    [tex]\nabla^2 \frac{1}{|x-x'|}=-4\pi \delta(x-x')[/tex]
     
  6. Nov 26, 2007 #5

    samalkhaiat

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    Last edited: Nov 26, 2007
  7. Nov 27, 2007 #6

    jmb

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    OK samalkhaiat I'll have a shot at your "remaining 8 marks", let me know if I'm missing something!

    From the divergence theorem we have,

    [tex]I = \int_{V} \nabla^2 (1/r) dV = \int_{S} \nabla (1/r) \cdot \mathbf{dS}[/tex]

    Since the RHS is a surface integral we don't need to worry about evaluating [tex]\nabla (1/r)[/tex] at the origin and so, choosing V as a sphere centred on the origin, we can straightforwardly write the surface integral as:

    [tex]I = \int_{0}^{2\pi} \int_{0}^{\pi} \frac{\partial}{\partial r} \left(\frac{1}{r}\right) r^2 \sin{\theta} \; d\theta \; d\phi = -4\pi[/tex]

    (since [tex]\frac{\partial}{\partial r} \left(\frac{1}{r}\right) = -\frac{1}{r^2}[/tex]) as required.
     
  8. Nov 27, 2007 #7

    marcusl

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    Jackson has similar expressions on p. 39 (he calls the integral C instead of I).
     
  9. Nov 27, 2007 #8

    samalkhaiat

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  10. Nov 28, 2007 #9

    jmb

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  11. Nov 29, 2007 #10

    samalkhaiat

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    Last edited: Nov 29, 2007
  12. Nov 30, 2007 #11

    jmb

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    Cool. Many thanks again to both marcusl and samalkhaiat for your help!
     
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